1

Analysis and Design Analysis and Design for Torsionfor Torsion

Dr. Husam Al Qablan

for Torsionfor Torsion

Introduction

A moment acting about a longitudinal axis of the member is called a torque, twisting moment or torsional moment, T.Torsion may arise as the

lt f

Dr. Husam Al Qablan

result of:(a)Primary or equilibrium

torsion: occurs when the external load has no alternative to being resisted but by torsion. Examples: curved girders and the three structures shown in Figure.

Introduction

secondary or compatibility torsion: in statically indeterminate structures from the requirements of continuity. the stressing of adjacent members as the beam twists permit a

Dr. Husam Al Qablan

as the beam twists permit a redistribution of forces to these members and reduces the torque that must be supported by the beam.

Examples of torsion

1. Floor systems: compatibility torque (perimeter beams supporting one or two way slab systems).

Dr. Husam Al Qablan

2. Floor system: equilibrium torque (circular beams).

3. Circular tie beams in mosques.

Torsional effects in reinforced concrete

Dr. Husam Al Qablan Dr. Husam Al QablanCompatibility Torsion

2

Dr. Husam Al Qablan

Shearing Stresses Due to Torsion in Un-cracked Members

Dr. Husam Al Qablan

Behavior of Circular Sections

Although circular sections are rarely a consideration in normal concrete construction, a brief discussion serves as a good introduction to the torsional behavior of other types of sections.The basic assumptions are:

1. Plane sections perpendicular to the axis of a circular member remains plane after torque is applied

Dr. Husam Al Qablan

plane after torque is applied.2. Radii of section stay straight (without warping).As a result of applying the torsion shearing stresses are set up on cross sections perpendicular to the axis of the bar as shown in Fig.

Behavior of Circular Sections

Shear stress is equal to shear strain times the shear modulus in the elastic range. If r is the radius of the element, J = πr4 /2 its polar moment of inertia,

Dr. Husam Al Qablan

J πr /2 its polar moment of inertia, and τmax is the maximum elastic shearing stress due to elastic twisting moment T, then from basic strength (mechanics) of material courses

maxrT=

Jτ

Behavior of rectangular sections

Such sections do not fall under the assumptions stated before. They warp when a torque is applied and radii don't stay straight. As a result axial as well as circumferential shearing stresses are generated. For a rectangular member, the corner elements do not distort at all (τcorners=0) and the maximum shear stresses occur at the midpoints of the long sides as

Dr. Husam Al Qablan

shown in Figure. These complications plus the fact that reinforcedconcrete sections are neither homogeneous nor isotropic make it difficult to develop exact mathematical formulations based on the physical models.

Dr. Husam Al Qablan

3

Dr. Husam Al Qablan

Hollow membersConsider a thin-wall tube subjected to a torsion T as shown in the Fig. below. If the thickness of the tube is not constant and varies along the perimeters of the tube, then equilibrium of an element like that shown in Figure b requires:

2AB 1 1 2 2 1 1 2CD= V dx = dx = qV t t t tττ τ τ→ → =

Dr. Husam Al Qablan

Where q is referred to as the shear flow and is constant.

Hollow membersIn order to relate the shear flow q to the torque T, consider an element of length ds as shown. This element is subjected to a force qds and

PT = rqdx∫

Dr. Husam Al Qablan

but rds = twice the area of the shadedtriangle, then

where Ao is the area enclosed by the middle of the wall of the tube. From the above equation τmax occurs where t is the least.

P q∫

2 oo

q TT = q =A t 2A tτ→ =

Principal stresses due to torsion

Principal tensile stresses eventually cause cracking that spirals around the body, as Shown by the line A-B-C-D-E I i f d h k ld

Dr. Husam Al Qablan

In reinforced concrete such a crack would Cause failure unless it was crossed by reinforcement. This generally takes the formof longitudinal bars in the corners and closedstirrups.

Principal stresses due totorsion and shear

The two shear stresses components add on oneside face (front side) and counteract each otheron the other. As the result inclined cracking starts on AB and extends across the flexural

Dr. Husam Al Qablan

tensile face. If bending moments are large, the cracks will extend almost vertically across theback face. The flexural compression zone near the bottom prevents the cracks from extending full height.

Behavior of RC members subjected to torsionWhen a concrete member is loaded inpure torsion, shear stresses develop. One or more cracks (inclined) developwhen the maximum principal tensile stress reaches the tensile strength of conc ete The onset of c acking

Dr. Husam Al Qablan

of concrete. The onset of cracking causes failure of an unreinforced Member. Furthermore the addition of longitudinal steel without

stirrups has little effect on the strength of a beam loaded in pure torsion because it is effective only in resisting the longitudinal component of the diagonal tension forces.

A rectangular beam with longitudinal bars in the corners and closed stirrups can resist increased load after cracking as shown in figure.

4

Behavior of RC members subjected to torsionAfter the cracking of a reinforced beam, failure may occur in several ways. The stirrups, or longitudinal reinforcement, or both, may yield, or, for beams that are over-reinforced in torsion, the concrete between the inclined cracks may be crushed by the principal compression stresses prior to yield

Dr. Husam Al Qablan

y p p p p yof the steel. The more ductile behavior results when both reinforcements yield prior to crushing of the concrete. Figure shows that ultimate strength of rc beams were the same for solid and hollow beams having the same reinforcement.

Space Truss AnalogyTheory

Assumptions:1. Both solid and hollow members

are considered as tubes.2. After cracking the tube is

idealized as a hollow truss

Dr. Husam Al Qablan

consisting of closed stirrups, longitudinal bars in the corners, and compression diagonals approximately centered on the stirrups. The diagonals are idealized as being between the cracks that are at angle θ, generally taken as 45 degrees for RC.

The cracking pure torsionKnowing that the principal tensile stress equal to the shear stress for elements subjected to pure shear, thus the concrete will crack when the shear stress equal to the tensile capacity of cross section. If we use conservatively 0.333√fc as tensile strength of concrete in biaxial tension-compression, and remembering that Ao must be some fraction of the area enclosed by the outside perimeter of the full concrete cross section Acp. Also, the value of t can, in general, be approximated as a fraction of the

ti A /P h P i th i t f th ti Th i

o

q T=t 2A t

τ =

Dr. Husam Al Qablan

ratio Acp/Pcp, where Pcp is the perimeter of the cross section. Then, assuming a value of Ao approximately equal to 2 Acp /3, and a value of t=3 Acp/4Pcp. Using these values in Eq. above yields:

2

3cpc

crcp

AfT =

p

According to ACI-R11.6.1

)(2;:

43;

32

o

cpcp

cp

cpcpo

tandAofvaluesngsubstituti

yxPxyAwhere

PA

tAA

+==

==

Dr. Husam Al Qablan

4

31 2

'

cru

cp

cpccr

TT

whenTorsionNeglect

PA

fT

φ≤

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=⇒

Vertical Stirrups and Longitudinal steel Reinforcement

Steel area in one leg stirrup leg is

oyv

ut Af

sTA

φ2=

Dr. Husam Al Qablan

oyvfφ

Longitudinal steel reinforcement

yo

hul fA

PTAφ2

=

Combined Shear and Torsion

dbV

w=τ

AT

2=τ

Shear Stress :

Torsional Stress :

Dr. Husam Al Qablan

tAo2

hohoho pAtAA /;85.0 ==For cracked section :

5

Dr. Husam Al Qablan

27.1 oh

h

wtv

ATp

dbV

+=+= τττ2

2

2

7.1 ⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=

oh

h

w ATp

dbVτ

ACI Requirement for Torsional Design

Equilibrium Torsion: Design for full uT Compatibility Torsion: reduce uT to the following

Nonprestressed member without a ial force

Dr. Husam Al Qablan

Nonprestressed member without axial force

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

cp

cpc P

Af

2'

31φ

Nonprestressed member with axial force

'

2'

33.01

31

cg

u

cp

cpc

fA

NPA

f +⎟⎟

⎠

⎞

⎜⎜

⎝

⎛φ

Neglect Torsion effect if the factored torsional moment is less than Nonprestressed member without axial force

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ cpc P

Af

2'083.0 φ

Dr. Husam Al Qablan

⎠⎝ cpP

Nonprestressed member with axial force

'

2'

33.01083.0

cg

u

cp

cpc

fA

NPA

f +⎟⎟

⎠

⎞

⎜⎜

⎝

⎛φ

The cross sectional dimensions shall be such that For solid section

⎟⎟⎠

⎞⎜⎜⎝

⎛+≤⎟

⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛= '

2

2

2

max 66.07.1 c

w

c

oh

hu

w

u fdb

VApT

dbV

φτ

Dr. Husam Al Qablan

For hollow section

⎟⎟⎠

⎞⎜⎜⎝

⎛+≤+= '

2max 66.07.1 c

w

c

oh

hu

w

u fdb

VApT

dbV

φτ

If NOT increase section dimensions

Reinforcement for torsion Recall ACI Eq (11-21)

oot TA

Dr. Husam Al Qablan

oo

oyv

utAfT

sA 6030;

cot2≤≤= θ

θφ

Combined shear and torsion reinforcement

sA

sA

sA tvtv 2

+=+

Maximum spacing of torsion reinforcement

⎪⎨⎧ p

ofsmallersh

8

Dr. Husam Al Qablan

⎪⎩⎨=

mmofsmallers

3008max

Spacing is limited to ensure the development of the ultimate torsional strength of the beam, to prevent excessive loss of torsional stiffness after cracking, and to control crack widths.

6

Minimum area of closed stirrups

( )⎪⎪

⎪⎪

⎨

⎧

=+wc

yv

w

tv sbf

fsb

oferlAA '062.0

35.0

arg2

Dr. Husam Al Qablan

⎪⎩ yvf

Minimum area of longitudinal torsional reinforcement

yv

wt

y

vth

t

y

cpcl

fb

sAwhere

ffp

sA

fAf

A

175.0

42.0 '

min,

≥

⎟⎠⎞

⎜⎝⎛−=

The longitudinal reinforcement shall be distributed around the perimeter of the closed stirrups with a maximum spacing of 300 mm

Dr. Husam Al Qablan

mm The longitudinal bars shall be inside the stirrups The longitudinal bars Shall have a diameter at least 0.042 times the stirrups spacing but not less than Ø10

Design Procedure for Combined Shear, Torsion and Moment

Step 1. Calculate the factored bending moment diagram or envelope for the member.

Step 2. Select b, d, h and As based on Mu. Note: for problem involving torsion, square cross-sections are

Dr. Husam Al Qablan

p ob e o g to s o , squa e c oss sect o s a epreferable.

Step 3. Given b and h, draw final Mu , Vu and Tu diagrams or envelopes. Calculate the reinforcement required for flexure.

Step 4. Determine whether torsion must be considered. Torsion must be considered if Tu exceeds the torque given by

⎟⎟⎞

⎜⎜⎛ cpA

f2

'0830 φ

Design Procedure for Combined Shear, Torsion and Moment

Dr. Husam Al Qablan

Otherwise, it can be neglected.

Where φ=0.75

⎟⎟

⎠⎜⎜

⎝ cpc P

f083.0 φ

Design Procedure for Combined Shear, Torsion and Moment

Step 5: If the torsion is compatibility torsion, the maximum factored torque may be reduced to

2

3cpc

ucp

AfT

pφ

≤

Dr. Husam Al Qablan

At sections d from the faces of the supports (the moments and shears in the other members must be adjusted accordingly). Equilibrium torsion cannot be adjusted.

cp

Step 6: Check shear stresses in the section under combined torsion and shear, for solid section

⎟⎟⎞

⎜⎜⎛

+≤⎟⎟⎞

⎜⎜⎛

+⎟⎟⎞

⎜⎜⎛

= '2

2

2

max 66.0 cchuu f

VpTVφτ

Dr. Husam Al Qablan

The critical section for shear and torsion is located a distance d from the face of the support.

⎟⎠

⎜⎝⎟

⎠⎜⎝

⎟⎠

⎜⎝

2max 7.1 cwohw

fdbAdb

φ

7

Design Procedure for Combined Shear, Torsion and Moment

Steps 7-9. Calculate the required transverse reinforcement for torsion and shear:

Compute Vs = Vu /Φ - Vc; then calc. Av/s=Vs/fyt dwhere fyt ≤4,20MPa.

Dr. Husam Al Qablan

If increase the size of the

cross section.

23 cs w> dV bf

Design Procedure for Combined Shear, Torsion and Moment

Find:

Then combine shear and torsional transverse reinforcement for a typical two-leg stirrup as:

t n

yto

A T=s 2 fA

Dr. Husam Al Qablan

reinforcement for a typical two leg stirrup as:

Check minimum transverse reinforcement requirements:

2v t v tA A A =s s s

+ +

2 0.062 , (0.35 )v t yt ytw wcf s/ or s/b bA A f f+ ≥

Solve for the required spacing of closed stirrups s, and compare it with ph/8 or 30cm maximum spacing for torsion (ACI 11.5.6.1) and d/2 or d/4 maximum spacing for shear.

Dr. Husam Al Qablan

Design Procedure for Combined Shear …Step 10: Compute longitudinal area of steel using the larger

of:

S ti f th i ( h ld t d 30 ) d b i

2cott yt hl

y

pA fA s f

θ=

,min5 0.175,

12c cp ytt t w

l hy y y

f A A A bfpA s s tf f f= − ⋅⋅⋅ ≥

Dr. Husam Al Qablan

Satisfy the spacing (should not exceed 30cm), and bar size requirements (the diameter of longitudinal bar may not be less than s/24 or 10mm). Torsion reinforcement must be symmetrically distributed around all cross section and that part which needs to be placed where As is needed must be added to As found in step 1. Torsion reinforcement must be extended at least a distance d+bt beyond the section where

2

12cpc

ucp

AfT

p≤

Additional remarks

1. Fy ≤420MPa to limit crack widths ACI 11.5.3.42. The transverse stirrup used for torsional

reinforcement must be of a closed form. The concrete outside the reinforcing cage is not well

Dr. Husam Al Qablan

co c ete outs de t e e o c g cage s ot eanchored, and the shaded region will spall off if the compression in the outer shell is large as shown in figure:

Additional remarksThus ACI 11.5.4.2 (a) requires that stirrups or ties must be anchored with a 135o hooks around longitudinal bars

Dr. Husam Al Qablan

if the corner can spall. ACI 11.5.4.2 (b) allows the use of a 90 degrees standard hook if the concrete surrounding the anchorage is restrained against spalling by a flange or a slab.

8

Additional remarks

3. If flanges are included in the computation of torsional strength for T and L-shaped beams, closed torsional stirrups must be provided in the flanges as shown in Figure.

Dr. Husam Al Qablan

Example: Torsion

wtownunfactoredkNPkNP

MpaffMpaf

LL

DL

yvy

c

+⎭⎬⎫

=

=

==

=′

8585

42020

Dr. Husam Al Qablan

Assume

Dr. Husam Al Qablan

mkNP

mkNwt

mkNwt

ult

factored

/238856.1852.1

/91.676.52.1

/76.5246.04.0

=×+×=

=×=

=××=

Dr. Husam Al Qablan

Moments at the support

mkNT

mkNM

/8.352/15.00365.115.0238

/5.3302/5.15.191.635.10365.135.1238

=×+×=

=××+×+×=

Dr. Husam Al Qablan

Design for Flexure

Assume mmda 75.1334/ ==

( ) ( )

,

26

3.1154002085.0

4207.186785.0

7.1867102/75.1335354209.0

5.3302/

mmbf

fAa

mmadf

MA

c

ys

y

us

=××

×==

=×−×

=−Φ

=

Dr. Husam Al Qablan

( )

2

26

1831

min

max

1831102/3.1155354209.0

5.330

mmA

Aforcheck

Aforcheck

forcheck

mmA

s

s

s

t

s

=

=×−×

=

ε

9

Should Torsion be Considered?

( )

Af

mmP

mmA

cpc

cp

cp

⎟⎞

⎜⎛⎟

⎞⎜⎛

=+=

=×=

24000020

20006004002

240000600400

622,

2

Dr. Husam Al Qablan

consideredbemustTorsionT

mkNP

fT

u

cp

cpc

⇒>=

=×⎟⎟⎠

⎜⎜⎝

×=⎟⎟

⎠⎜⎜

⎝Φ= −

05.88.35

.05.8102000

240000122075.0

126

θ

The torsion is needed for equilibrium

⇒ Design for 8.35=uT

Is the section large enough to resist torsion?

⎟⎟⎠

⎞⎜⎜⎝

⎛+≤⎟

⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛= '

2

2

2

max 66.07.1 c

w

c

oh

hu

w

u fdb

VApT

dbV

φτ  

1

Dr. Husam Al Qablan

dbfV wcc'

61

=

ohA = area within centerline of closed stirrups

Assume 40mm cover mm10Φ

Dr. Husam Al Qablan

( ) mmyxPmmymmx

mmxyA

h

oh

1640)25101040260031010402400

158100310510 2

=+==−×−==−×−=

=×==

MPadb

MPa

MPa

w795220660

2061

75081

8.11581007.1

1640108.35535400107.245

2

2

623

=⎟⎟⎞

⎜⎜⎛

+≤

=⎟⎟⎠

⎞⎜⎜⎝

⎛

×

××+⎟

⎟⎠

⎞⎜⎜⎝

⎛

××

⇒

Dr. Husam Al Qablan

MPadb

MPaw

795.22066.075.08.1 =⎟⎟⎟

⎠⎜⎜⎜

⎝

+≤

The cross section is large enough

Compute the stirrups area required for shear

( )scu VVV +Φ≤

kNdbfV wcc 5.159105354002011 3' =×××== −

Dr. Husam Al Qablan

mmmms

A

dfV

sA

ors

dfAV

kNV

f

v

y

svyvs

s

wcc

/748.0535420101.168

1.1685.15975.0

7.24566

23

=××

=⇒

==

=−=

For shear ⇒ require stirrups with mmmms

Av /748.0 2=

Compute the stirrup area required for torsion

TA

mmNT

oont

n

6030

.107.4775.0

108.35 66

≤≤

×=×

=

θ

Dr. Husam Al Qablan

mmmms

A

mmAA

Afs

t

ooho

oo

oyv

nt

/422.04201343852

107.47

45

13438515810085.085.0

6030;cot2

26

2

=××

×=

=

=×=×=

≤≤=

θ

θθ

10

Add the stirrup area and select stirrups

mmmms

AsA

sA

sA

tv

tvtv

/59.1422.02748.0

2

2=×+=

+=

+

+

Ch k i i ti

Dr. Husam Al Qablan

Check minimum stirrups

( )

⎪⎪⎩

⎪⎪⎨

⎧

=+

yv

w'c

yv

w

tv

fbf

161

fb

31

oferargls

A2A

OK

mm/mm266.0420

40020161

mm/mm317.0420400

31

sA

min

2

2

tv

⇒

⎪⎪

⎩

⎪⎪

⎨

⎧

=×

=

⇒ +

Dr. Husam Al Qablan

Use ⇒10Φ Area for two legs = 2mm157

mm7.98s59.1s

15759.1s

A tv =⇒=⇒=+

Use ⇒12Φ Area for two legs = 2mm226

mm2.142s =

Check maxs

OKmm300

mm2058

16408p

ofsmallersh

max ⇒⎪⎩

⎪⎨⎧ ===

⇒ Use 12Φ closed stirrups at 125 mm on center

Design the longitudinal reinforced for torsion

2yvfA ⎟⎞

⎜⎛

⎟⎞

⎜⎛

Dr. Husam Al Qablan

22l

t

2

yl

yvh

tl

mm69245cot4204201640422.0A

422.0s

A

cotff

ps

AA

=⎟⎠⎞

⎜⎝⎛××=

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛= θ

yl

yvh

t

y

cp'c

min,l ff

psA

fAf

125A ⎟

⎠⎞

⎜⎝⎛−=

OKmmmmAl ⇒<=××−×

= 22min, 692372

4204201640422.0

42024000020

125 we need 6

bars < 30 mm spacing

Each area 2mm33.1156/692A =≥

Dr. Husam Al Qablan

The min bar dim {

⎪⎩

⎪⎨⎧

≥

=×=

10

25.5125042.0

φ

mmspacingstirrup

Provide 144φ in the bottom half of the beam and add

22

mm76414*4692 =⎟⎟

⎠

⎞⎜⎜⎝

⎛−

π to the flexural steel

2mm1907761831A +

Dr. Husam Al Qablan

s mm1907761831A =+=

Example: Compatibility Torsion

The one-way joist system shown in the figure below supports a

total factored dead load of 2/5.7 mkN and factored live load of 2/8 mkN , totaling 2/5.15 mkN . Design the end span AB, of the

exterior spandrel beam on the grid line 1 the factored dead load

Dr. Husam Al Qablan

exterior spandrel beam on the grid line 1. the factored dead load of the beam and the factored loads applied directly to it total

mkN /16 . The spans and loading are such that the moments and the shears can be calculated by using the moment coefficients from ACI section 8.3.3. Use Mpafc 30=′ ,. Mpaff yvy 420==

11

Dr. Husam Al Qablan

Compute the bending moments for the beam.

In laying out the floor, it was found that joist with an overall depth of 470 mm would be required. The slab thickness is 110 mm. the spandrel beam was made the same depth, to save forming costs. The columns

Dr. Husam Al Qablan

supporting the beams are 600 mm square. For simplicity in forming the joist, the beam overhangs the inside face of the columns by 50 mm. thus, the initial choice of the beam size is h = 470 mm, b = 650 mm, and d = 405 mm.

The joist reaction per meter of length of beam is

mkNmmkNwln /1.722

30.9/5.152

2=

×=

Th t t l l d th b i

Dr. Husam Al Qablan

The total load on the beam is

mkNw /1.88161.72 =+=

Using mmln 6600= . The moments in the edge beam are as follows

Exterior end negative mkNwlM nu /9.239

16

2−==

Midspan positive mkNwlM nu /1.274

14

2==

First interior negative mkNwlM nu /8.383

10

2−==

The areas of steel required for flexure are as follows:

Dr. Husam Al Qablan

Exterior end negative: 21791mmAs =

Midspan positive: 22046mmAs =

First interior negative: 22865mmAs =

The actual steel will be chosen when the longitudinal

torsion reinforcement has been calculated.

Compute the final moment shear and

Dr. Husam Al Qablan

moment, shear, and torsion diagrams

The exterior negative moment in the joist

mkNwlM nu /9.55

243.95.15

24

22−=

×==

Although this is a bending moment in the joist,

it acts as a twisting moment on the edge beam.

Dr. Husam Al Qablan

g g

As shown in the figure below, this moment and

the shear of 72.1 kN/m act at the face of the edge beam.

The torque at the center of the beam = t = 79.3 kN.m/m

81.5kN.m/m the torque transferred to the col

12

If the two ends of the beam A-B are fixed

against rotation by the column,

the total torque at each end will be

mkNtlT n .7.2612

6.63.792

=×

==

The total shear at end A of the beam will be

66188 ×

Dr. Husam Al Qablan

kNdV

kNV

u

u

2552

6.61.88@

7.2902

6.61.88

=×

=

=×

=

The total shear at end B of the beam will be

kNdV

kNV

u

u

7.2982

6.61.88@

3.3342

6.61.8815.1

=×

=

=×

×=

Dr. Husam Al Qablan

Dr. Husam Al Qablan Dr. Husam Al Qablan

Should torsion be considered?

consideredbemustTorsionT

mkNPAf

T

mmP

mmA

u

cp

cpc

cp

cp

⇒>=

=×⎟⎟⎠

⎞⎜⎜⎝

⎛×=⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛Φ=

=+++++=

=×+×=

−

8.137.261

.8.13102960

345100123075.0

12

29601010110360360650470

345100360110650470

622,

2

θ

Dr. Husam Al Qablan

Equilibrium or compatibility torsion?

The torque resulting from the 25-mm offset of the axes

of the beam and column is necessary for equilibrium torque.

The torque at the ends of the beam due to this is

mkN .3.726.6025.01.88 =××⇒

On the other hand, the torque resulting from

the moments at the ends of the joists exist only

because the joint is monolithic and the edge beam

has a torsional stiffness. If the torsional stiffness

were to decease to zero, this torque would disappear.

Dr. Husam Al Qablan

This part of the torque is therefore compatibility torsion.

mkN

PAf

dTcp

cpcu

.1.55102960

34510033075.0

3@

62

2,

=×⎟⎟⎠

⎞⎜⎜⎝

⎛×=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛Φ=⇒

−

But not less than the equilibrium torque

13

Assuming the remaining torque after redistribution

is evenly distributed along the length of the spandrel

beam. The distribution reduced torque t, due to moments

at the ends of the joists has decreased to

mkNt .192/)405.026.6(

1.55=

×−=⇒

Dr. Husam Al Qablan

Adjust the moments in the joists

Because the analysis procedure (ACI moment

coefficients) assumed an exterior support (spandrel

beam) with zero torsional stiffness,

No load or moment redistribution would be required.

Is the section large enough to resist torsion?

⎟⎟⎠

⎞⎜⎜⎝

⎛+≤⎟

⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛= '

2

2

2

max 66.07.1 c

w

c

oh

hu

w

u fdb

VApT

dbV

φτ

dbfV wcc'

61

=

mmp

mmAoh

1868)577377(2

209989577377 2

=+=

=×=

Dr. Husam Al Qablan

mmph 1868)577377(2 =+=

MPadb

dbMPa

MPa

w

w423.33066.0

3061

75.08.1

781.12099897.1

1868101.55405650107.298

2

2

623

=⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+≤

=⎟⎟⎠

⎞⎜⎜⎝

⎛

×

××+⎟

⎟⎠

⎞⎜⎜⎝

⎛

××

⇒

The cross section is large enough

Compute the stirrups area required for shear in the edge beam

kNdbfV wcc 313.240104056503061

61 3' =×××== −

At the left of the beam (End B)

2076.1405420

24031375.0

103.334/

3

=×

−×

=−

==df

VVdf

Vs

Ay

cu

y

sv φ

At d f d B

Dr. Husam Al Qablan

At d from end B

9286.0405420

24031375.0

107.298/

3

=×

−×

=−

==df

VVdf

Vs

Ay

cu

y

sv φ

Compute the stirrups required for torsion

42020998985.0210/;

cot2/ 6

××××

==φ

θφ u

oyv

ut TAf

Ts

A

At end B, 5583.0.8.62 =⇒=s

AmkNT tu

At d from end B, 4902.0.1.55 =⇒=s

AmkNT tu

At d from end A, 4902.0.1.55 =⇒=s

AmkNT tu

Dr. Husam Al Qablan

Add the stirrup area and select stirrups

909.14902.029286.0

2

=×+=

+=

+

+

sA

sA

sA

sA

tv

tvtv

Provide No. 13M closed stirrups

End A: one @75 mm, seven @150 mm

End B: one @75 mm, 12@125 mm, then @200 mm

on centers through the rest of the span

Design the longitudinal reinforcement for torsion

2cotff

ps

AA yvh

tl ⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛= θ

Dr. Husam Al Qablan

29161118684902.0 mmA

fs

l

yl

=×××=

⎟⎠

⎜⎝⎠⎝

2min,

,

min,

960

118684902.042012345100305

125

mmA

ff

ps

AfAf

A

l

yl

yvh

t

yl

cpcl

=

××−××

=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−=

Use 2960mmAl =

To satisfy ACI requirements, we need 3 bars

at the top and bottom and one halfway up each side

22 1208/960/ mmmmbarAs ==

Exterior end negative moment

2215112031791 mmA =×+=

Dr. Husam Al Qablan

215112031791 mmAs =×+=

Use 8 No. 19M

First interior negative moment

2322512032865 mmAs =×+=

Use 7 No. 25M

14

Midspan positive moment

2240612032046 mmAs =×+=

Use 5 No. 25M

Dr. Husam Al Qablan Dr. Husam Al Qablan

Dr. Husam Al Qablan Dr. Husam Al Qablan